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GREYUIT [131]
3 years ago
11

The attractive electrostatic force between the point charges 6.14×10−6 c and q has a magnitude of 0.845 n when the separation be

tween the charges is 9.26 m . part a find the sign and magnitude of the charge q.
Physics
1 answer:
Nataly [62]3 years ago
3 0
Coulomb's law states that F = k(q1)(q2)/r^2, where k = 9 x 10^9 N-m^2/C^2
Substituting the given values:

0.845 N = (9x10^9 N-m^2/C^2)(6.14x10^-6 C)(q) / (9.26^2)
q = 0.00131 = 1.31 x 10^-3 C

Since the force is attractive, the point charges must be of opposite signs, and q is negatively charged (-1.31 x 10^-3 C).
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What kind of force will move an object from its rest resting ? A. inertial B. balanced C. unbalanced D. along a straight line
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4 years ago
A 1.1 kg hammer strikes a nail. Before the impact, the hammer is moving at 4.5 m/s; after the impact it is moving at 1.5 m/s in
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Answer:

132 N

Explanation:

Given that a  1.1 kg hammer strikes a nail. Before the impact, the hammer is moving at 4.5 m/s; after the impact it is moving at 1.5 m/s in the opposite direction. If the hammer is in contact with the nail for 0.025 s, what is the magnitude of the average force exerted by the hammer on the nail

From Newton 2nd law of motion,

Change in momentum = impulse.

Change in momentum = m( V - U )

Substitute all the parameters into the formula

Change in momentum = 1.1 ( 4.5 - 1.5 )

Change in momentum = 1.1 × 3

Change in momentum = 3.3 kgm/s

Impulse = Ft

That is,

Ft = 3.3

Substitute time t into the formula above

F × 0.025 = 3.3

F = 3.3 / 0.025

F = 132 N

Therefore, the magnitude of the average force exerted by the hammer on the nail is 132 N.

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Explanation :

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